91Ó°ÊÓ

To rationalize the denominator, multiply both the numerator and the denominator by the conjugate of the denominator, which is \((\sqrt{x}+\sqrt{y})\). This will eliminate the root in the denominator: $$ \frac{x-y}{\sqrt{x}-\sqrt{y}} \times \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}} $$

Now, we will multiply the numerators together and the denominators together: $$ \frac{(x - y)(\sqrt{x} + \sqrt{y})}{(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})} $$ Let's apply the difference of squares formula for the denominator and apply the distributive property for the numerator: $$ \frac{x\sqrt{x} + xy - y\sqrt{x} - y\sqrt{y}}{x - y} $$ Now, we can see that there are common terms in the numerator, so let's group them together: $$\frac{x (\sqrt{x} - \sqrt{y}) + y (\sqrt{x} - \sqrt{y})}{x - y}$$ Now factor out the common term \((\sqrt{x} - \sqrt{y})\) from the numerator: $$\frac{(\sqrt{x} - \sqrt{y})(x + y)}{x - y}$$

Now that the given expression has been simplified, we can see if it's equal to the target expression \((\sqrt{x} + \sqrt{y})\): $$ \frac{(\sqrt{x} - \sqrt{y})(x + y)}{x - y} = \sqrt{x}+\sqrt{y} $$ To verify this, set \(z = \sqrt{x} - \sqrt{y}\). Then, we can rewrite the given expression as: $$ \frac{z(x + y)}{x - y} $$ Making this substitution, we can rewrite the equation as follows: $$ \frac{z(x-y)}{x-y}=z $$ Since the expression is only valid when \(x \neq y\), the denominator is never equal to zero, meaning we can cancel out the \((x-y)\) terms. Thus, our simplified expression is \(z\). Finally, substitute back the original values for z: $$ \sqrt{x} - \sqrt{y} = \sqrt{x}+\sqrt{y} $$ Now we have shown that the original expression is equivalent to \(\sqrt{x}+\sqrt{y}\).